WUCT121 Discrete Mathematics Graphs Tutorial Exercises Solutions

WUCT121
Discrete Mathematics
Graphs
Tutorial Exercises
Solutions
WUCT121 Graphs: Tutorial Exercise Solutions 1

Graphs
Question1 Let G = { V , E}
be a graph. Draw a graph with the following specified
properties, or explain why no such graph exists.
(a) A graph having V = { v1,
v2
, v3,
v4
, v5},
; E = { e1 , e2
, e3,
e4
, e5}
, where
e 1 = ( v1,
v3
) , e 2 = ( v2
, v4
) , e 3 = ( v1,
v4
) , e 4 = ( v3,
v3
) and e 5 = ( v3,
v5
) .
How many components does G have
v 1
v 2
e 1
e 3
e 2
e 4
v 3
e 5
v 5
v 4
G has one component.
(b) A connected graph having V = { v1,
v2
, v3,
v4
, v5},
and ∀vi
∈V
, δ ( vi
) = 2 .
How many edges does G have
v 1
e 1
v 2
v 5
e 5
e 4
v 4
e 3
e 2
v 3
G has 5 edges
WUCT121 Graphs: Tutorial Exercise Solutions 2

Question2 Either draw a graph with the following specified properties, or explain
why no such graph exists:
(a) A graph with four vertices having the degrees of its vertices 1, 2, 3 and 4.
v 1
e 1
v 2
e 5
e 2
e 3
v 4
v 3
e 4
(b) A simple graph with five vertices with degrees 2, 3, 3, 3, and 5.
It is impossible to draw this graph. A simple graph has no parallel edges nor any
loops. There are only 5 vertices, so each vertex can only be joined to at most four
other vertices, so the maximum degree of any vertex would be 4. Hence, you
can’t have a vertex of degree 5.
(c) A simple graph in which each vertex has degree 3 and which has exactly 6 edges.
v 1
e 1
v 2
e 4
e 5
e 2
e 6
v 4
e 3
v 3
WUCT121 Graphs: Tutorial Exercise Solutions 3

(d) A graph with four vertices having the degrees of its vertices 1, 1, 2 and 2.
v 1
e 1
v 2
e 2
e 3
v 4
v 3
(e) A graph with four vertices having the degrees of its vertices 1, 1, 2 and 6.
v 1
e 1
v 2
e 2
v 4
e 3
v 3
e 4
e 5
(f) A graph with five edges having the degrees of its vertices 1, 1, 3 and 3.
Not possible, number of edges is 5, thus the sum of the degrees of the vertices
4
must be 2 × 5 = 10 ∑δ
( v i ) = 1+
1+
3 + 3 = 8 ≠ 2×
5
i=
1
WUCT121 Graphs: Tutorial Exercise Solutions 4

Question3 Let G = { V , E}
be the graph drawn below.
e 1 e 2
v 1
e 5
v 2
e 3
e 4
v 3
(a) Draw a sub-graph H = { V H , EH
} of G for which ∑δ ( v)
= 4 .
v V
2 possibilities are:
∈ H
e 1
e 2
v 1
e 2
v 1
e 5
v 2
There are other possibilities.
(b) Draw a sub-graph H = { V H , EH
} of G having 2 components, one vertex of even
degree and one pair of vertices of odd degree
2 possibilities are:
e 1
v 1
v 2
e 3
v 3
WUCT121 Graphs: Tutorial Exercise Solutions 5

e 2
v 1
v 2
e 4
v 3
There are other possibilities.
Question4
(c) Either write down a path with the specified properties, or explain why no such
path exists.
(i) A path of length 5 from v 1 to v 2 .
One possible: v 1 , e1
, v1,
e2
, v1,
e1
, v1,
e2
, v1,
e5
, v2
(ii) A path of length 3 from v 1 to v 3.
One possible: v 1 , e2
, v1,
e5
, v2
, e3,
v3
(a) Draw all spanning trees for K 3 .
v 1
K 3
e 3
e 1
v 3 e 2 v 2
v 1
v 1
v 1
e 1
e 3
e 3
e 1
v 3 e 2 v 2
v 3 e 2 v 2
v 3 v 2
WUCT121 Graphs: Tutorial Exercise Solutions 6

(b) Draw 8 different spanning trees for K 4 . It can be shown that there are
spanning trees for
K n . Can you write down 16 spanning trees for K 4
n−2
n
K 4
v 1
e 4
v 4
e 5
e 6
e 1
e 3
v 2
e 2
v 3
v 1
e 1
v 2
v 1
v 2
v 1
e 1
v 2
v 1
e 1
v 2
e 2
e 4
e 2
e 4
e 4
e 2
v 4
v 1
v 4
e 6
e 3
e 1
e 3
v 3
v 2
v 3
v 4
v 1
v 4
e 5
e 3
e 1
e 3
v 3
v 2
v 3
v 4
v 1
e 4
v 4
e 6
e 3
v 3
v 2
e 2
v 3
v 4
v 1
e 4
v 4
e 5
v 3
v 2
e 2
v 3
v 1
v 4
v 1
e 4
v 4
e 5
e 6
e 5
e 1
e 1
v 2
v 3
v 2
v 3
v 1
v 4
v 1
v 4
e 5
e 6
e 6
e 3
e 1
v 2
v 3
v 2
e 2
v 3
v 1
e 4
v 4
v 1
v 4
e 5
e 6
e 5
e 3
v 2
v 3
v 2
e 2
v 3
v 1
v 4
v 1
e 4
v 4
e 5
e 6
e 6
e 3
v 2
e 2
v 3
v 2
v 3
WUCT121 Graphs: Tutorial Exercise Solutions 7

(c) Use Kruskal’s algorithm to find the minimum spanning tree for the following
weighted graph. Write down the number of steps you need to complete the
algorithm
1
v 1
2
v 2
2
1
3
v 3
3
v 4
Edge Weight Will adding
edge make
circuit
1 ( v1,v2
)
2 ( v2,v4
)
3 ( v1,v3
)
4 ( v2,v3)
5 ( v3,v4
)
6 ( v1,v4
)
Action taken
Cumulative
Weight of
Subgraph
e = 1 No Added 1
e = 1 No Added 2
e = 2 No Added 4
e = 2 Yes Not added 4
e = 3 Yes Not added 4
e = 3 Yes Not added 4
Minimum spanning tree is
v 1
v 2
v 3
v 4
The algorithm can be completed in 3 steps.
WUCT121 Graphs: Tutorial Exercise Solutions 8

WUCT121 Graphs: Tutorial Exercise Solutions 9